Question

If polynomial 2x 4 + 3x 3 + 2kx ​2 +3x+6 is exactly divisible by (x+2) then value of k is: Could you plz tell it in detail fast.

Answer 1

p (x)=x+2
p (x)=0
0=x+2
-2=x
we put -2 in the place of x
p (x)= 2x 4+ 3x 3+2kx2+3x+6
p (x)=0
0= (2×-2×4)+(3×-2×3)+(2×k×-2×2)+(3×-2)+6
0=-16+(-18)+(-8k)+(-6)+6
0=-16-18-8k-6+6
0=-34-8k
0+34=8k
34=8k
34/8=k
17/4=k

Answer 2

Answer:

k = 1

Step-by-step explanation:

Polynomial : $2x^4 + 3x^3 + 2kx^2+3x+6$

Since (x+2) divides the polynomial completely

So,

$p (x)=x+2\\p (x)=0\\So, x+2=0\\x=-2$

We put -2 in the place of x

$2x^4 + 3x^3 + 2kx^2+3x+6=0$

$2(-2)^4 + 3(-2)^3 + 2k(-2)^2+3(-2)+6=0$

$16-24+ 8k-6+6=0$

$8k=8$

k = 1

Hence the value of k is 1